Question

Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 2 mole of H2(g) with 2 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol. H2(g)+Cl2(g)→2HCl(g)

Answer 1

Given :

Hydrogen gas,$H_2$ , reacts explosively with gaseous chlorine, $Cl_2$, to form hydrogen chloride, HCl(g).

The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

$H_2(g)+Cl_2(g) - >2HCl(g)$

To Find :

The enthalpy change for the reaction of 2 mole of $H_2(g)$ with 2 mole of $Cl_2(g)$ if both the reactants and products are at standard state conditions .

Solution :

We know ,

$\Delta H^o_{rxn}=\Delta H^{o}f_{products}-\Delta H^of_{reactants}\\\\\Delta H^o_{rxn}=2\times (-92.3)-(0+0)\\\\\Delta H^o_{rxn}=-184.6\ kJ/mol$

( Since , $H_2$ and $Cl_2$ are their most elementary form . So their enthalpy of formation is 0 kJ )

Therefore , the enthalpy change for the reaction of 2 mole of $H_2(g)$ with 2 mole of $Cl_2(g)$ if both the reactants and products are at standard state conditions is -184.6 kJ/mol .

Hence , this is the required solution .

Answer 2

Answer:

-184.6 KJ/mole

Explanation:

H2(g) + Cl2(g) ------------> 2HCl(aq)    

Δ H0 rxn = Δ H0f products - Δ H0f reactants

= 2×-92.3 - (0 + 0)

= -184.6KJ/mole

H2(g) + Cl2(g) ------------> 2HCl(aq)       Δ H0rxn   = -184.6KJ/mole